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Math Help - Show that x is an accumulation point

  1. #1
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    Show that x is an accumulation point

    Let { x_n}, \ n \geq 1, be a bounded sequence of real numbers and let x := sup{ x_n | \ n \geq 1}.
    If for all n in N , x_n < x, prove that x is an accumulation point of { x_n}

    This question is given me alot of trouble, anyone can help solve this plz?
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  2. #2
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    To show that it is an accumulation point you need to show that for every interval containing x, the interval also contains some terms from the sequence different from x.

    Mathematically you need to show for every epsilon, [x-\epsilon,x+\epsilon] contains s_n for some n and s_n\neq x.

    Hint: you need to use the fact that supremum is the least upperbound
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  3. #3
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    can it be proven using sub-sequences?
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  4. #4
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    Quote Originally Posted by Cato View Post
    can it be proven using sub-sequences?
    I am not sure what you mean by that?
    So what is the accumulation point definition given in your book?
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  5. #5
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    In the book, it says,

    A point s, is called an accumulation point of a sequence { s_n} if there exists a sub-sequence of { s_n} that converges to s.
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  6. #6
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    Quote Originally Posted by Cato View Post
    Let { x_n}, \ n \geq 1, be a bounded sequence of real numbers and let x := sup{ x_n | \ n \geq 1}.
    If for all n in N , x_n < x, prove that x is an accumulation point of { x_n}

    This question is given me alot of trouble, anyone can help solve this plz?
    In the case where x_{n}{\color{red}{\leq}}x is true for all n\in\mathbb{N}, I dont believe that x defined as in your definition is not always an accumulation point for all bounded sequences.

    Counter Example
    . Let x_{n}:=\frac{2n}{n^{2}+1} for n\in\mathbb{N}.
    It is obvious that x_{n} is bounded.
    It is easy to see that x=1 (this value holds with n=1) and x_{n}<1 for all n\in\mathbb{N}\backslash \{1\}.
    But x_{n} tends to 0 as n tends to \infty. \rule{0.3cm}{0.3cm}

    I feel that the claim is true when x_{n}{\color{red}{<}}x holds for all n\in\mathbb{N}, I will be thinking on it.
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  7. #7
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    Quote Originally Posted by Cato View Post
    In the book, it says,

    A point s, is called an accumulation point of a sequence { s_n} if there exists a sub-sequence of { s_n} that converges to s.
    Ok

    Since x is the supremum, by definition if we substract a little bit from x, then x is no longer a lower bound.

    So for every n we have that x-\frac{1}{n} is not a lower bound, i.e, there are terms in the sequence which are in (x-1/1, x). Choose such element with the least index, say x_{n_1}
    Choose x_{n_2} in (x-1/2, x) with the same properties.
    If you continue, then you have a subsequence. Then you need to prove that this subsequence converges to x too.
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  8. #8
    Senior Member bkarpuz's Avatar
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    Here is the proof you want.

    Proof.
    Let \{x_{n}\} be a bounded sequence and x_{n}<x hold for all n\in\mathbb{N}, where x:=\sup\{x_{n}:n\in\mathbb{N}\}.
    On the contrary, suppose that x is not an accumulation point for \{x_{n}\}.
    In this case, we may find a sufficiently small \varepsilon>0 such that x_{n}\not\in(x-\varepsilon,x+\varepsilon) for all n\in\mathbb{N} or equivalently x_{n}\leq x-\varepsilon for all n\in\mathbb{N}.
    In this present case, we arrive at the following contradiction:
    x=\sup\{x_{n}:n\in\mathbb{N}\}\leq\sup\{x-\varepsilon:n\in\mathbb{N}\}=x-\varepsilon.
    The proof is hence completed. \rule{0.3cm}{0.3cm}
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  9. #9
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    thanks alot guys, you are both very helpful. Much appreciated.
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